More answers

radar

7 years ago

\[\sqrt{x ^{4}+7}\] or\[\sqrt{x ^{4+7}}\]??

anonymous

7 years ago

na I need a specific answer

anonymous

7 years ago

it is apparently factorable

anonymous

7 years ago

There is no good way to simplify that problem. It is simply \[\sqrt(x^4+7)\]
Owlfred, your reply was useless.
Helpneededlady, 11 is prime, and that was not this persons question anyway.
Jman, is there more to the problem?

anonymous

7 years ago

no

anonymous

7 years ago

thats it I have to simplify it further

anonymous

7 years ago

Find f(√x + 2) if f(x) = x4 + 5

anonymous

7 years ago

that is the problem

anonymous

7 years ago

only it is x to the fourth

anonymous

7 years ago

Well then you already gave us the solution to your problem, and it cannot be simplified further.

anonymous

7 years ago

no it has to be simplified

anonymous

7 years ago

I did notice some calculation errors, that could be your problem.
The answer ends up being \[(\sqrt(x+2))^4+5\]. At this point you can simplify and get a different answer.

anonymous

7 years ago

f(√x + 2) if f(x) = x4 + 5
(√x + 2)^4 + 5
= (x+2)^2 +5
= x^2 + 2x +9

anonymous

7 years ago

i dont think we r allowed to do that suzi

anonymous

7 years ago

why?
f(√x + 2) means x = √x + 2
then plug into equation

anonymous

7 years ago

suzi, (sqrtx + 2)^4 is not (x + 2)^2

anonymous

7 years ago

Guyc is correct, my bad I didn't notice the calculation errors, but her general procedure was correct.